Difference between revisions of "2014 AMC 12A Problems/Problem 23"
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and we can see that for each <math>n+m=k</math> there are <math>k-1</math> <math>(n,m)</math> combinations so the above sum is equivalent to: | and we can see that for each <math>n+m=k</math> there are <math>k-1</math> <math>(n,m)</math> combinations so the above sum is equivalent to: | ||
<cmath>\sum^{\infty}_{k=2}\dfrac{k-1}{10^{-2k}}</cmath> | <cmath>\sum^{\infty}_{k=2}\dfrac{k-1}{10^{-2k}}</cmath> | ||
− | we note that the sequence | + | we note that the sequence starts repeating at <math>k = 102</math> |
yet consider <cmath>\sum^{101}_{k=99}\dfrac{k-1}{10^{-2k}}=\dfrac{98}{10^{10^{198}}}+\dfrac{99}{10^{10^{200}}}+\dfrac{100}{10^{10^{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)</cmath> | yet consider <cmath>\sum^{101}_{k=99}\dfrac{k-1}{10^{-2k}}=\dfrac{98}{10^{10^{198}}}+\dfrac{99}{10^{10^{200}}}+\dfrac{100}{10^{10^{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)</cmath> | ||
so the decimal will go from 1 to 99 skipping the number 98 | so the decimal will go from 1 to 99 skipping the number 98 | ||
and we can easily compute the sum of the digits from 0 to 99 to be <cmath>45*10*2=900</cmath> subtracting the sum of the digits of 98 which is 17 we get | and we can easily compute the sum of the digits from 0 to 99 to be <cmath>45*10*2=900</cmath> subtracting the sum of the digits of 98 which is 17 we get | ||
<cmath>900-17=883\textbf{(B) }\qquad</cmath> | <cmath>900-17=883\textbf{(B) }\qquad</cmath> |
Revision as of 22:20, 7 February 2014
Problem
The fraction where
is the length of the period of the repeating decimal expansion. What is the sum
?
Solution
the fraction can be written as
.
similarly the fraction
can be written as
which is equivalent to
and we can see that for each
there are
combinations so the above sum is equivalent to:
we note that the sequence starts repeating at
yet consider
so the decimal will go from 1 to 99 skipping the number 98
and we can easily compute the sum of the digits from 0 to 99 to be
subtracting the sum of the digits of 98 which is 17 we get